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\bf{Question}

\quad Find the evolute of each curve $\gamma$, both as (i) the
locus of centres of curvature and (ii) the envelope of the
normals:

\begin{tabular}{ll} \hspace{2cm}(a)&\quad
$\gamma(t)=(t-\sin t,1-\cos t)$\\ \hspace{2cm}(b)&\quad
$\gamma(t)=(2\cos t+\cos2t,2\sin t-\sin2t).$
\end{tabular}



\bf{Answer}

\begin{description}
\item{(a)}

\begin{eqnarray*}
\gamma(t) & = & (t - \sin t, 1 - \cos t)\\ \dot{\gamma}(t) & = &
(1 - \cos t, \sin t)\\ \ddot{\gamma}(t) & = & (\sin t, \cos t)\\
K(t) & = & -1/4\sin \frac{t}{2}
\end{eqnarray*}

Now
\begin{eqnarray*}
N(t) & = & (-\sin t, 1 - \cos t)/(\sin^2 t + (1 \cos t)^2
)^{\frac{1}{2}}\\ & = & (2-2\cos t)^{-\frac{1}{2}} (-\sin t,
1-\cos t)\\ & = & \frac{1}{2 \sin \frac{t}{2}} ( -\sin t, 1 -\cos
t)
\end{eqnarray*}

Hence centre of curvature is
\begin{eqnarray*}
\gamma(t) + (K(t))^{-1} N(t) & = & ( t- \sin t, 1 - \sin t)\\ & =
& 2 ( - \sin t, 1- \cos t)\\ & = & (t + \sin t, -1 + \cos t)
\end{eqnarray*}

OR: Equation of normal at $\gamma(t)$ is
$$x(1-c)+ys=(t-s)(1-c)+(1-c)S = t(1-c)$$ Differentiation with
respect to $t$ gives $$xs + yc = 1-c+ts$$ Solving for $x,y$ gives
$$(x,y) = (t + \sin t, -1 + \cos t)$$

(Observe that the evolute is also a cycloid: put $t=\pi+u$, and
translate by $(\pi, -2)$.)

\item{(b)}

\begin{eqnarray*}
\gamma (t) & = & (2 \cos t + \cos 2t, 2 \sin t - \sin 2t), \ \
\textrm{hypocycloid}\\ \dot{\gamma}(t) & = & (-2 \sin t- 2\sin 2t,
2 \cos t - 2 \cos 2t)\\ \ddot{\gamma}(t) & = & 2(-\cos t - 2\cos
2t, -\sin t + 2\sin 2t)\\ & &\\ \frac{1}{4} \| \dot{\gamma} \|^2 &
= & (\sin t + \sin 2t)^2 + (\cos t - \cos 2t)^2\\ & = & 2 + 2(\sin
t \sin 2t - \cos t \cos 2t)\\ & = & 2 -2 \cos 3t = 4 \sin^2
\frac{3t}{2}
\end{eqnarray*}

\begin{eqnarray*}
\frac{1}{2}(\dot{x} \ddot{y} - \dot{y} \ddot{x} ) & = &  (-\sin t
- \sin 2t)(-\sin t + 2 \sin 2t)\\ & &  + (\cos t - \cos 2t)(\cos t
+ 2 \cos 2t)\\ & = & 1 -2 - \sin t \sin 2t + \cos t \cos 2t\\ & =
& -(1-\cos 3t) = -2 \sin^2 \frac{3t}{2}\\ \Rightarrow K(t) & = &
-4 \sin^2\frac{3t}{2}/ \left (4 \sin \frac{3t}{2} \right )^3\\ & =
& -1/16\sin\frac{3t}{2}
\end{eqnarray*}

Now $$N(t) = \frac{1}{4 \sin \frac{3t}{2}} (-\cos t + \cos 2t,
-\sin t - \sin 2t),$$ so centre of curvature is
\begin{eqnarray*}
\gamma(t) + (K(t))^{-1} N(t) & = & (2 \cos t + \cos 2t, 2 \sin t -
\sin 2t)\\ & &  - 4(-\cos t + \cos 2t, - \sin t - \sin 2t)\\ & = &
3(2 \cos t - \cos 2t, 2 \sin t + \sin 2t)
\end{eqnarray*}
Which we see is another hypocycloid, $=-3 \times \gamma (t +
\pi)$.

OR: Equation of normal at $\gamma(t)$ is $$-x(\sin t + \sin 2t) +
y(\cos t - \cos 2t) = \ \cdots \ =-3\sin 3t$$ Differentiate with
respect to $t$ to give $$-x(\cos t + 2\cos 2t) + y(-\sin t +2\sin
2t) = -9\cos 3t$$ 'Solving' for $x,y$ gives (eventually) $$(x,y) =
3(2\cos t - \cos 2t, 2 \sin t + \sin 2t)$$.

\end{description}



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